The integration by parts that you mention in a comment went bad for a truly minor reason. But I have seen versions of this slip before, so it is maybe worth commenting on.

We want $\int_0^b 2te^{-t}\,dt$. Let $u=2t$, and $dv=e^{-t}\,dt$. You decided to find an antiderivative of $2te^{-t}$.
We get
$$\int 2te^{-t}\,dt=-2te^{-t}+\int 2e^{-t}\,dt=-2te^{-t}-2e^{-t}+C=-2(t+1)e^{-t}+C. \qquad(\ast)$$
This seems to be precisely what you did, apart from the $+C$ that I added because of excessive fussiness.

We now want to "plug in $b$, take away the result of plugging in $0$." But because we are so accustomed to the result of plugging in $0$ being $0$, it is all too easy not to see the $0$. However, in this case, and often with integration of exponentials, the important action is at $0$. We find that
$$ \int_0^b 2te^{-t}\,dt=2-2(b+1)e^{-b}.$$
The rest is routine limit taking.

Remark: As a parenthetical remark, I would prefer to work with the definite integral, as in
$$\int_0^b 2te^{-t}\,dt=\left.(-2te^{-t})\right|_0^b+\int_0^b 2e^{-t}\,dt.$$
Less algebra, and the first part dies at both ends.

Isn't the LP too much here, if the OP is struggling with integration by parts? (I think your answer is OK, but maybe you can simply calculate the integral with $x=y^2$)
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Pedro Tamaroff♦Mar 22 '12 at 19:52